Given the following chemical reaction, A → B. If the concentration of A is doubled the rate increases by a factor of 2.83, what is the order of the reaction with respect to A?
A
1
B
0.5
C
1.5
D
0
E
2
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Write the general rate law for the reaction: \(\text{rate} = k [A]^n\), where \(k\) is the rate constant, \([A]\) is the concentration of A, and \(n\) is the order of the reaction with respect to A.
Express the change in rate when the concentration of A is doubled: \(\frac{\text{rate}_2}{\text{rate}_1} = \frac{k (2[A])^n}{k [A]^n} = 2^n\).
Use the given information that the rate increases by a factor of 2.83 when the concentration of A is doubled, so set \(2^n = 2.83\).
Take the logarithm of both sides to solve for \(n\): \(n = \frac{\log(2.83)}{\log(2)}\).
Calculate the value of \(n\) using the logarithms to find the order of the reaction with respect to A.